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Solutions

General Aptitude

1. Ans. A. Hurtful would be best option as person is complaining about her. 2. Ans. B.

3. Ans. B. Given 40% of deaths on city roads are drunken driving

w.k.t in pie chart 100% →360° 1% →360

100

40% →

360

100

× 40 40% →144°

4. Ans. D. Let H is house hold consumption and P is the other

consumption.

Given H×0.8+P×1.7 =( H+P)×0.75 Ratio is negative. 5. Ans. A. Past Tense is used.

6. Ans. D. Option D is correct according to the passage 7. Ans. A. No. of sub groups such that every sub group has at least one Indian

=7+9+9+3+6+9+9+3+3+3+1=56. Alternate method Sub groups containing only Indians

= 1 2 3

3 3 3 3 3 1 7C C C

Subgroups containing one Indian and rest chinese

= 1 1 2 3

3 3 3 3 3[3 3 1] 21C C C C

Sub groups containing two Indian and remaining Chinese

2 1 2 33 3 3 3 21C C C C

Sub groups containing three Indian and remaining

Chinese

3 1 2 33 3 3 3 7C C C C

Total no. of sub group = 7 + 21 + 21 + 7 = 56

8. Ans. C. Down- up-Down

9. Ans. A. Given speeds both car & Truck = 36 km/hour They travel in 1 hr = 36 km = 36000 m.

∴ Maximum no. of vehicles than can use the bridge in I

hour =36000

50

m

m =720×2=1440 vehicles

Alternate method Length of truck + gap required = 10+20 = 30m

Length of car + gap required = 5+15 = 20m Alternative pairs of Truck and car needs 30+ 20 = 50 m. Let 'n' be the number of repetition of (Truck + car) in 1 hour (3600 sec). Given speed 36km /hr 10m /sec

5036 /

3600sec

m nkm hr

50/ sec 10 / sec

3600

nm m

36000720( )

50n Truck car

So, 720 Truck car passes 720 2 1440vehicles.

10. Ans. A. Following circular seating arrangement can be drawn.

Only one such arrangement can be drawn. The person on third to the left of V is X.

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Electronics & Communications 1. Ans. C.

fclock = 5MHz; Tclock = 0.2 × 610

sec

Texecution = 1.4μs

No. of T-state required = 1.4

0.2 = 7

2. Ans. A.

For an input-output relation if the present output depends on present and past input values then the given system is “Causal”. For the given relation,

For n ranging from 0 to 10 present output depends on

present input only. At all other points present output depends on present and

past input values. Thus the system is “Causal”. Stability If x[n] is bounded for the given finite range of n i.e. 0 < n 10 y[n] is also bounded. Similarly x[n]–x [n–1] is also bounded at all other values of n

Thus, the system is “stable”. 3. Ans. B. y1 = 1, y 2 = x, y 3 = x2 Consider

2

1 2 3

1 2 3

1 2 3

1

' ' ' 0 1 2

" " " 0 0 2

y y y x x

y y y x

y y y

= 1

20 1

x =2 ≠0

y1, y2, y3 are linearly independent x

4. Ans. C.

A=

1 2 3 4 5

5 1 2 3 4

4 5 1 2 3

3 4 5 1 2

2 3 4 5 1

For eigen values ( ), |A- I|=0

1 2 3 4 5

5 1 2 3 4

04 5 1 2 3

3 4 5 1 2

2 3 4 5 1

1 1 2 3 4 5R R R R R R

15 15 15 15 15

5 1 2 3 4

04 5 1 2 3

3 4 5 1 2

2 3 4 5 1

(15

1 1 1 1 1

5 1 2 3 4

04 5 1 2 3

3 4 5 1 2

2 3 4 5 1

15 =0

=15

5. Ans. A.

Given 1(0.1 40)j m

Here 0.1 P

m

We know that,

1 8.686 0.1 0.8686P PdB dB

m m m m

6. Ans. A.

Silicon atoms act as P- type dopants in Arsenic sites and

n- type dopants in Gallium sites.

7. Ans. C.

|M |=

5 10 10

1 0 2

3 6 6

= 5(0-12) – 10(6-6) = -60-0+60=0

But a 2×2 minor, 5 10

1 0

=0-10=-10≠0

Rank = 2

8. Ans. C.

As per the change carrier profile, base – to – emitter

junction is reverse bias and base to collector junction is forward bias, so it works in Inverse active.

9. Ans. D.

Miller effect increase input capacitance, so that there will

be decrease in gain in the high frequency cutoff frequency.

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10. Ans. A.

Dutyde of o/p = 2 5 100 30%

CLK CLK

CLK

T T

T

11. Ans. A.

In phase lag compensator pole is near to jω- axis,

12. Ans. B.

Unequal propagation delay

Case I :

Gate 1 →2ns

Gate 2 →1ns

Case II :

Gate 1→1nsec

Gate 2 → 2nsec

∴ Either x = 1, y = 0 or x = 0, y =1

13. Ans. A.

Required probability

= 6 1 1 1

6 6 6

=

1

36=0.028

14. Ans. A.

If x(t) = -x(-t) the given periodic signal is odd symmetric. For an odd symmetric signal an for all n

If x(t)=

0

x t

0

0 2

T

where T0 is

fundamental period then the given condition satisfies half-wave symmetry.

For half-wave, symmetrical signal all coefficients an and

bn are zero for even value of n.

15. Ans. A.

G(s)=( 2)

1)

( 3)

(ps s

s

s

If p=1, ess (for ramp input) = 6

kv =

1

6

p=1 , ess (for ramp input) = 0

kp = ∞, ess =

1

1 pk = 0

16. Ans. D.

Vbi = VT l n

2

1N

N

=0.25ln

18

15

10

10

= 0.173V.

17. Ans. B.

Vsat = 15V, -Vsat = -15V

VUTP = (15 3) 5 2

3 3 715 3

1V

VLTP= 3 3 6 3 315

( 15 3) 5 1

3

8V

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18. Ans. A.

V1 =1

1 02 j

V1 = 1

2

100tan

24

V2= 1

100 02

j

j

V2 =

21 1

2

11 tan tan

24

V2-V1=4

-1 1 1tan tan tan

2 2 4

tan 14

rad/sec

19. Ans. B.

For ISI free pulse, If P(t) is having spectrum P(f)

Then )( S

k

P f kR

= constant

R S = 2 KSpa

Thin condition is met by pulse given in option B.

20. Ans. A.

G (s)=1

1

1

111

1

p

p

q p q

p

S S

S q S a S

b b

If q pS

=3S , when p=0 and q=3, then

It has -60dB/dec at ω=∞

21. Ans. C.

A good trans conductance amplifier should have high input and output resistance.

22. Ans. A.

For two independent random variable

I (X; Y ) H (X) = H (X Y )

H(X Y)= H(X) for independent X and Y

I (X; Y)= 0

23. Ans. D.

If a system is non-causal then a pole on right half of the

s-plane can give BIBO stable system. But for a causal

system to be BIBO all poles must lie on left half of the complex plane.

24. Ans. D.

DPCM Block diagram

e q [n] is quantized e[n]

e[n]is difference of message

signal sample with its prediction.

25. Ans. C.

C= Blog2

0

1S

N B

Where S= 24

t t erPG A

r

= 2 2 2 2

4 .

( ) 4. ( )

t er e er et

PA A t A A tp

r r

=2 2

. .t er eP A A tS

A r

Channel capacity remain same.

26. Ans. A.

Let f(x) = 𝑥3 + 𝑥 − 1 f’(x) = 3𝑥2 + 1

Given X0 =1

By Newton Raphson method.

1st iteration ,

x1 = x0 - ( )

'( )

f

f

0

0

x

x = 1-

(1)

'(1)

f

f =1-

1

4 =

3

4 =0.75

2nd iteration. X2 = x1 - 1

1

( )

'( )

f

f

x

x= 0.75 -

(0.75)

'(0.75)

f

f-

0.17

2.69

=0.69

27. Ans. B.

u0(t) =5cos(2000𝛑t)

F0=10kHz

U1(t)=5cos(22000𝛑t)

f1==11kHz

For u0(t) and U1(t) to be orthogonal, it is necessary that

f1-f0 =2

n

T ; (11-10) ×

310 =1

2T

T = 3

1

2 10 = 0.5 m sec

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28. Ans. A.

Av =

1

V

V

0 =

c

e

R

r

re = T

E

V

I

VG =12 47

120

= 4.7 V

VG = VEE + IERE

IE = 3

4.7 0.7

2 10

= 2mA

re =

25

2 =12.5Ω

AV =

3 3|| 2 10 || 8 10128

12.5

c

e

LR R

r

29. Ans. A.

V1=4 3

100 04 3 5 12

j

j j

V1=4 3

100 09 9

j

j

V2=5 12

100 04 3 5 12

j

j j

5 12100 0

9 9

j

j

2

1V

V

=

2 2

2 2

5 12 132.6

54 3

5 12

4 3

j

j

30. Ans. D.

2( 1)dy

x ydx

------- (1)

Put x+y-1 = t

1+dy dt

dx dx

1+dy dt

dx dx -1

From (1), dt

dx-1 =

2t

21dt

tdx

2

1

1dt dx

t

1tan ( )t x C

1tan ( 1)x y x C

31. Ans. A.

G(s)=

2

2

( 2 2)

( 3 2)

K s s

s s

C.E = 1 + G (s) = s 2 - 3s + 2 + ks 2 + 2ks + 2k=0

= s 2 (1 + k) + s (2k – s) + 2k + 2 = 0

If closed loop system to be stable all coefficients to

positive k >-1

1 1.5 1k k k

So, k >1.5

32. Ans. A.

The straight line joining A(0, 2, 1) and B(4, 1, -1) is

0 2 1

4 0 1 2 1 1

x y z

2 1

4 1 2

x y zt

(say)

x = 4t, y = 2 - t, z = 1 - 2t

dx = 4dt, dy=-dt, dz=-2dt

For x = 0 t = 0

For x = 4 t =1

I= (2 2 2 )C

zdx ydy xdz

=

1

0

2(1 2 )4 2(2 )( ) 2(4 )( 2 )t

t dt t dt t dt

=

1 2

0

30( 30 4) 4

2t

tt dt t

33. Ans. A.

C=A

W

C∝1

w and W ∝

1

doping

C∝ doping

1

2C

C =

1

2doping

doping =

16

14

10

10 = 100 = 10

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34. Ans. A. For circular polarization, the phase difference between Ex & Ey is π/ 2

The phase difference for linear polarization should be π

So the wave must travel a minimum distance such

that the extra phase difference of π/ 2 must occur.

βyllmin –βxlmin =2

lmin

min

0

2/ 2 / 2 y x y x

ln n n

cn

lmin =

620

2

1.5 10 1.510

4 0.0001 44

0.375 10 0.3

7

5

y xn

m cm

n

35. Ans. A.

Given x[n]=1,2,1; h[0]=1 h[n]=1,a,b

Y[n]=1,2+a,2a+b+1,2b+a,b It is given that y[1] =3 ∴ 2+a=3 a=1

Similarly 2a+b+1=4 b=3-2(1)=1

b=1 ∴y[3]=2(1)+1=3

Y[4]=b=1 ∴10y[3]+y[4]=30+1=31

36. Ans. C.

F(s) =4 2 1 0s s

Let take 2s =t

2t +t+1=0

t=1 3

2

i

Where t=2s =

1 3

2

i

2s =1 3

2

i =

2

23

j

e s

1 3

2

i =

2

3

j

e

S=

2

6

j

e

and s=

2

6

j

e

Hence two roots contain RHS and two roots contain LHS plane.

37. Ans. A. vd =𝛍n∈

𝛍n = dv

=

7 2

7

1020

5 10 sec

cm

V

E=

4

5/

1 10

Vv cm

d

Jdrift=nqvd=nqμn∈

Jdrift = nqva= nqμnε = 1610 ×1.6×

1910×20×

410 =1.6

KA/c2m

38. Ans. A.

VG=8 5

5

=5V

VGS = VG- IDRs=5-310 ID

ID=1

2 μnCox

W

L

2( )GS TV V

ID=1

2×1×

310 2( )GS TV V

ID=1

2×

3102

35 10 1DI =

310

2

2

34 10 DI

ID =

310

2

6 2 316 10 8 10DI

0.5×310 2

DI -5ID-38 10 =0

ID=8mA,2mA ID must be least value So ID =2mA

39. Ans. A. Since DC components in same in Sx (f) and Sy(f)

[ ( )] [ ( )]E x t E y t

2( )E x t = Area under

Sx(f) = | | |2 2f fe df e df

2 ( )E y t = Area under

Sy(f)=

1

12

2

0

2 2 | 2[1 ]1

ff f e

e df e df e

2[ ( )]E x t ≠2[ ( )]E y t

2[ ( )]E y t ≠2

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40. Ans. A.

∆P=∆n= GLo21

L

L

τp=GLo× τp=1710 ×

1

2 ×

410=

1

2×

1310 /3cm

|Jp1 diff| =qDp

13

19 2

4

110

21.6 10 100 16 /0.1 10

2

dpA cm

dx

41. Ans. B.

G(s) =

2 2 2 ( 2)

K

s s s

C.E =3 2 76 4 0s s s k

If system to stable

24>k+4∩K+4>0 k>-4∩k<20 (i) Stable condition -4<k+20 Means If k =10 system stable k =100 system unstable

Or G(jω)=1 1

2

2tan tan

2 2

If ω 0 G(jω)= 04

k

ω G(jω)= 0 270

So If k =10 touching point= 0.5

If k =100 touching point =5

N= P-Z, Here P=0

N=-Z

If closed loop system to be stable, then z=0,N=0,

So, k=0 is stable system

42. Ans. A.

IB= 3

(2 0.7)0.108

12 10mA

IC= 3

(5 0.2)1

4.8 10mA

𝛃=1

9.2590.108

c

B

I

I

𝛂=9.259

0.9031 1 9.259

43. Ans. A.

PQ = 2 2x y is the radius of variable circle at some Z.

2 2 2 3pQ x y z (Given)

∴ Volume of region revolved around z-axis =

1 1 42 3 1

0

0 0

( ) | 0.794 4

zPQ dz z dz

44. Ans. C.

Given the direction of propagation is

ˆ ˆx ya a

The orientation of E field is ˆ ˆ ˆ2x y za a j a

The dot product between above two is =1-1+0=0

It is a plane wave

We observed that

ˆ ˆ ˆ ˆ ˆ, 2x y x y zP a a a a andj a are normal to each other.

So electric field can be resolved into two normal

component along ˆ ˆx ya a and j2 ˆ

xa

The magnitude are 2 and 2 and θ =2

So elliptical polarization.

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45. Ans. D. In given diagram

When Xin=0 2 State When Xin= 1 3 State 46. Ans. B.

Fourier series coefficient ak is unaffected by scaling operating. Thus (I) is true and (II) is false. T= 1sec for x(t) and if it compressed by ‘3’ then the resultant period

T=1

3

∴ Fundamental frequency =

1

2

T

=6π rad/sec.

Thus (III) is correct. 47. Ans. A.

48. Ans. C. We have series of f(x) around x=0 is f(x)=

2 3' "(0) (0) (0) ""(0)

2! 3!

x xf xf f f

(upto powers of „x‟ less than or equal to „3‟) Given

f(x) = 2

(0) 1x xe f

2' '( ) (1 2 ) (0) 1x xf x e x f

2 2'' ''( ) (1 2 ) 2 (0) 3x x x xf x e x e f

2 2 2''' 3

'''

( ) (1 2 ) 4(1 2 ) 2(1 2 )

(0) 7

x x x x x xf x e x e x x e

f

∴ f(x)=2x xe

= 1+x.1+

2 32 33 7

(3) (7) 12! 3! 2 6

x xx x x

49. Ans. B. MVI A, 33H A 33H

MVI B, 78H B 78H

ADD B B ABH

CMA A 54H

ANI 32H A 10H

A 0011 0011 A 1010 1011

B 0111 1000 B 0101 0100

_________

1010 1011 0101 0100 0011 0010 ________ 0001 0000

50. Ans. A.

Under dc condition inductor acts as short all

∴ Itotal=15

151

A

1 2( ) ( (0 ) ( ) ( )t

i t i i e i

(0 ) (0 ) 0i i A

3

2( ) (0 15) 15t

i t e

Itotal= 3

215

(1 )3 3

t talt

o eI

2=5

3

21t

e

t=0.34sec

51. Ans. A.

The null occurs along axis of the antenna which is θ=90°, ∅=45°

52. Ans. A. Given Amplifier is using –ve feed back

0

01f

AA

A

5

0

1; 10

80A

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5

5

0

1079.93

1 10 / 8

8 (1 ) 10008

f

cut

A

f Hz A Hz

2( )

1 ( / )

f

f

cut

AA

f f

=2

3

79.93

15 101

10008

=44.3

53. Ans. A.

It is given that,

H[0]=1 1 1

; [1] ; [2] &3 3 3h h

H[n]=0 for n<0 and n>2.

∴ h[n]=h[0]δ[n]+h[1]δ[n-1]+h[2]δ[n-2]

1[ [ ] [ 1] [ 2]]

3n n n

Apply DTFT on both sides,

∴ H(ω) =21

[1 ]3

j je e

Given that H(ω0) =0 & 0< ω0.<π

0 0( ) 0 &0H

0 0 03

2 2 20

1( ) 1 0

3

j j j

H e e e

∴1+2 03 je

cos

0 02

Consider |H 0 | cos0 1

2 2

∴ 0 =2

3

0 =2.094

54. Ans. B.

0 1 2 3 5 minF M M M M M term

55. Ans. A.

Given: (t)=

2sin(300 )0

600 0

tt

t

t

Thus h(t) =600sin c(300t)

∴H(f) = 2rect300

f

Given x(t) =4cos200πt+800cos400πt

In f-domain X(f)=2[δ(f-100)+δ(f+100)]+4[δ(f-200)+δ(f+200)]

***